text: Given an n x n square matrix , find sum of all sub | An efficient C ++ program to find sum of all subsquares of size k x k ; Size of given matrix ; A O ( n ^ 2 ) function to find sum of all sub - squares of size k x k in a given square matrix of size n x n ; k must be smaller than or equal to n ; 1 : PREPROCESSING To store sums of all strips of size k x 1 ; Go column by column ; Calculate sum of first k x 1 rectangle in this column ; Calculate sum of remaining rectangles ; 2 : CALCULATE SUM of Sub - Squares using stripSum [ ] [ ] ; Calculate and print sum of first subsquare in this row ; Calculate sum of remaining squares in current row by removing the leftmost strip of previous sub - square and adding a new strip ; Driver program to test above function
code: #include <iostream> NEW_LINE using namespace std ; #define n  5 NEW_LINE void printSumTricky ( int mat [ ] [ n ] , int k ) { if ( k > n ) return ; int stripSum [ n ] [ n ] ; for ( int j = 0 ; j < n ; j ++ ) { int sum = 0 ; for ( int i = 0 ; i < k ; i ++ ) sum += mat [ i ] [ j ] ; stripSum [ 0 ] [ j ] = sum ; for ( int i = 1 ; i < n - k + 1 ; i ++ ) { sum += ( mat [ i + k - 1 ] [ j ] - mat [ i - 1 ] [ j ] ) ; stripSum [ i ] [ j ] = sum ; } } for ( int i = 0 ; i < n - k + 1 ; i ++ ) { int sum = 0 ; for ( int j = 0 ; j < k ; j ++ ) sum += stripSum [ i ] [ j ] ; cout << sum << " ▁ " ; for ( int j = 1 ; j < n - k + 1 ; j ++ ) { sum += ( stripSum [ i ] [ j + k - 1 ] - stripSum [ i ] [ j - 1 ] ) ; cout << sum << " ▁ " ; } cout << endl ; } } int main ( ) { int mat [ n ] [ n ] = { { 1 , 1 , 1 , 1 , 1 } , { 2 , 2 , 2 , 2 , 2 } , { 3 , 3 , 3 , 3 , 3 } , { 4 , 4 , 4 , 4 , 4 } , { 5 , 5 , 5 , 5 , 5 } , } ; int k = 3 ; printSumTricky ( mat , k ) ; return 0 ; }